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 10-12-2012, 05:39 PM #1 Junior Member   Join Date: May 2011 Posts: 290 How to solve these math problems? Hi! Would be delighted if somebody explained how to solve following problems. 1. Single-color tokens of blue, red, green, or yellow are placed in a single line so that the pattern of blue, red, green, yellow, blue, red, green, yellow repeats throughout. If the first token in the line is blue, which of the following tokens is green? (A) 117th (B) 118th (C) 119th (D) 120th (E) 121st 2. The integers 1 through 6 appear on the six faces of a cube, one on each face. If three such cubes are rolled. what is the probability that the sum of the numbers on the top faces is 17 or 18. (A) 1/108 (B) 1/54 (C) 1/27 (D) 1/18 (E) 1/16 Thank you in advance! Reply
 10-12-2012, 06:53 PM #2 Senior Member   Join Date: Feb 2011 Location: Northern VA Posts: 1,021 1. If the pattern repeats "blue, red, green, yellow," then it moves in cycles of four. Listing the first few tokens helps you visualize the progression: B, R, G, Y, B, R, G, Y, B, R, G, Y... Here we see that the Y token always falls on a multiple of four, and the G token is just in front of every Y token in the order. Thus, the G token will always be at a position one less than a multiple of four. 120 is the only multiple of four provided, so 119 is the answer. 2. To find the probability in a dice problem, we have to find the number of desired possibilities (in this case, sums of 17 or 18) and divide that by the number of total possibilities (in this case, total number of sums possible). The total number of sums is 6 * 6 * 6 = 216. This is because we can add any of 6 possibilities of the first die, 6 in the second, and 6 in the third to my sum. Thus, 6 numbers * 6 numbers * 6 numbers = 216 sums. Now, how many desired possibilities are there? It should be clear that there is 1 possibility in which all the dice show a 6 and therefore sum to 18. Then, for the sum of 17, there are three possibilities: (5, 6, 6), (6, 5, 6), and (6, 6, 5). Hence, there are 4 desired possibilities. We're almost there! The probability, as I said earlier, is desired / total. 4/216 = 1/54, so that's your answer. Hope that helped! Reply
 10-13-2012, 11:01 AM #3 Junior Member   Join Date: May 2011 Posts: 290 Thank you very much! It helped a lot Reply
 10-13-2012, 02:07 PM #4 Junior Member   Join Date: May 2011 Posts: 290 I just got another problem I can't solve. Could anyone help? If 0 ≤ x ≤ y and ( x + y )^2 - ( x - y )^2 ≥ 25, what is the LEAST possible value of y ? Reply
 10-13-2012, 02:43 PM #5 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 Apply difference of two squares on the LHS: ((x+y) + (x-y))((x+y) - (x-y)) ≥ 25 (2x)(2y) ≥ 25 4xy ≥ 25 --> xy ≥ 25/4 Since y is at least x (x ≥ 0) , the minimum value of y occurs when y = x = 5/2. Reply
 10-13-2012, 05:53 PM #6 Junior Member   Join Date: May 2011 Posts: 290 Thank you! It seems so easy when I read an explanation, yet I had no clue what to do after I'd found that xy ≥ 25/4. But now I see. Reply

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